In a satellite moving round any planet, the gravitational force is effectively balanced. If an ice cube exists there, and it melts with passage of time, its shape will

remain unchanged
change to spherical
become oval-shaped with long-axis along the orbit plane
become oval-shaped with long axis perpendicular to orbit plane

Solution

Because of surface tension.

A capillary tube of radius r is immersed in a liquid. The liquid rises to a height h. The corresponding mass is m.What mass of water shall rise in the capillary if the radius of the tube is doubled?

Solution

A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from the pan of a balance, so that it comes in contact with the water in a glass vessel. If an extra 3.97. If is required to pull it away from water, the surface tension of water is

72 dyne cm^-1
70.80 dyne cm^-1
63.35 dyne cm^-1
60 dyne cm^-1

Solution

In case A,when an 80 kg skydiver falls with arms and legs fully extended to maximize his surface area, his terminal velocity is 60 m/s. In Case B, when the same skydiver falls with arms and legs pulled in and body angled downward to minimize his surface area, his terminal velocity increases to 80m/s.In going from Case A to Case B,which of the following statements most accurately describes what the skydiver experiences?

F_{air resistance} in creases and pressure P increases
F_{air resistance} in creases and pressure P decreases
F_{air resistance} decreases and pressure P increases
F_{air resistance} remains the same and pressure P increases

Solution

For the first part of the question, remember that terminal velocity means the acceleration experienced becomes zero.

Since a = 0 m/s², then,ΣF_y F_{air resistance} - F_W = 0

F_{air resistance} = F_W mg

For the second part of the question, while the velocity is higher,the acceleration is still zero. Therefore, the

F_{air resistance} still equal to the skydiver’s weight.

F_{air resistance Case A} = F_{air resistance Case B}

What has changed is the surface area of the skydiver.Since pressure is P = F/A,as A decreases, the pressure experienced in creases.

P_AA_A= P_B A_B= mg

Since A_A > A_B , then P_A < P_B

One drop of soap bubble of diameter D breaks into 27 drop shaving surface tension σ. The change in surface energy is

2πσD²
4πσD²
πσD²
8πσD²

Solution

Two capillary of length L and 2L and of radius Rand 2R are connected in series. The net rate of flow of fluid through them will be (given rate to the flow through single capillary, \(X=\frac{\pi PR^{4}}{8\eta L}\))

⁸⁄₉X
⁹⁄₈X
⁵⁄₇X
⁷⁄₅X

Solution

A body of density ρ’ is dropped from rest at a height h into a lake of density ρ where ρ > ρ’ neglecting all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface:

\(\frac{h}{\rho -\rho '}\)
\(\frac{h\rho '}{\rho}\)
\(\frac{h\rho '}{\rho-\rho '}\)
\(\frac{h\rho}{\rho-\rho '}\)

Solution

A tank is filled with water upto a height H. Water is allowed to come out of a hole Pin one of the walls at a depth h below the surface of water (see fig.) Express the horizontal distance X in terms H and h.

\(X=\sqrt{h(H-h)}\)
\(X=\sqrt{\frac{h}{2}(H-h)}\)
\(X=2\sqrt{h(H-h)}\)
\(X=4\sqrt{h(H-h)}\)

Solution

Two drops of the same radius are falling through air with asteady velocity of 5 cm per sec. If the two drops coalesce,the terminal velocity would be

10 cm per sec
2.5 cm per sec
5 × (4)^1/3 cm per sec
5 × √3 cm per sec

Solution

An open vessel containing water is given a constant acceleration a in the horizontal direction. Then the free surface of water gets sloped with the horizontal at an angle θ, given by

θ = cos^-1 ^g⁄_a
θ = tan^-1 ^a⁄_g
θ = sin^-1 ^a⁄_g
θ = tan^-1 ^g⁄_a

Solution

θ = tan^-1 ^g⁄_a

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